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3 years ago

How many moles of carbon are in the sample?

Chemistry

1 answer:

mafiozo [28]3 years ago

7 0

Answer:

12.01 grams per mole

Explanation:

theres no sample or example so if theres grams of Carbon, convert it to moles using 12.01g/mol

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A mixture initially contains AA, BB, and CC in the following concentrations: [A][A]A_1 = 0.550 MM , [B][B]B_1 = 1.40 MM , and [C

Alex787 [66]

Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc

Step 1: The balanced equation

A+2B<->C

Step 2: The initial concentrations

[A] = 0.550 M

[B]= 1.40 M

[C] = 0.600 M

Step 3: The concentraions at equilibrium

[A] = 0.550 -X = 0.430 M

[B]= 1.40 -2X M

[C] = 0.600 + X = 0.720 M

X = 0.120 M

[A] = 0.550 - 0.120 = 0.430 M

[B]= 1.40 -2*0.120 = 1.16 M

[C] = 0.600 + 0.120 = 0.720 M

Step 4: Calculate Kc

Kc = [C] / [A][B]²

Kc = 0.720 / (0.430*1.16²)

Kc = 1.244

The value of the equilibrium constant KC is 1.244

5 0

3 years ago

Which of the following is considered a greenhouse gas?

Sergeeva-Olga [200]

The primary greenhouse gases are water vapor, carbon dioxide, methane, nitrous oxide, and ozone. So, the correct answer among these options is water vapor, that is considered to be a greenhouse gas.

4 0

2 years ago

A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to

melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the Henderson-Hasselbach (H-H) equation:

pH = pKa + log $\frac{[CH_3COO^-]}{[CH_3COOH]}$

For acetic acid, pKa = 4.75.

We calculate the original number of moles for acetic acid and acetate, using the given concentrations and volume:

CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻

CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to calculate the new pH, by using the new concentrations:

pH = 4.75 + log $\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L}$ = 5.54

6 0

3 years ago

Polymer formed from methylene diphenyl isocyanate

marin [14]

Answer:

Polymeric MDI is a mixture of. monomeric MDI as well as larger molecular weight oligomers of MDI, and is a brownish. liquid at room temperature and may have a slight odor. Commercial MDI products are. often mixtures of monomeric and polymeric MDI and can contain other additives as well.

Explanation:

5 0

3 years ago

What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r

saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

Explanation:

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$$\frac{1500}{2658}=0.564 \text { moles }$ of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol

= 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

= 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0

3 years ago